Two point charges 5×10^-8 culomb and …

Two charges q_A = 5 {tex}\times{/tex} 10^-8C and q_B = -3 {tex}\times{/tex} 10^-8C
Distance between two charges, r = 16 cm = 0.16 cm
Consider a point O on the line joining two charges where the electric potential is zero due to two charges.

From the figure we can see that, x = distance of point O from charge q_A
Electric potential at point O due to q_A,
{tex}\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{q}_{\mathrm{A}}}{4 \pi \varepsilon_{0}(\mathrm{AO})}{/tex}
{tex}=9 \times 10^{9} \times \frac{5 \times 10^{-8}}{\mathrm{x}}{/tex}
{tex}=\frac{450}{\mathrm{x}}{/tex}
Electric potential at point O due to q_B
{tex}\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{q}_{\mathrm{B}}}{4 \pi \varepsilon_{0}(\mathrm{BO})}{/tex}
{tex}=9 \times 10^{9} \times \frac{-3 \times 10^{-8}}{0.16-\mathrm{x}}{/tex}
{tex}=\frac{-270}{0.16-X}{/tex}
Since the total electric potential at O is zero,
{tex}\Rightarrow{/tex} V_A + V_B = 0
{tex}\Rightarrow \frac{450}{x}+\left(-\frac{270}{0.16-x}\right)=0{/tex}
{tex}\Rightarrow \frac{450}{x}=\frac{270}{0.16-x}{/tex}
{tex}\Rightarrow \frac{5}{x}=\frac{3}{0.16-x}{/tex}
On cross multiplying we get,
5 {tex}\times{/tex} (0.16 - x) = 3x
{tex}\Rightarrow{/tex} = 0.8 - 5x = 3x
{tex}\Rightarrow{/tex} 8x = 0.8
{tex}\Rightarrow{/tex} x = 0.1m = 10cm (from charge q_A)
{tex}\therefore{/tex} at a distance of 10cm from the positive charge, the potential is zero between the two charges.